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Even without calculations it is clear that ( \(\sqrt{2}\) , \(\frac{31}{3}\) ) cannot lie on the line \(y = 5x + 3\) . If x is irrational, y must also be irrational.

What do they mean by \((\sqrt{2}, \frac{31}{3})\) is irrational?

Which of the following points is not on the line \(y = 5x + 3\) ?

a \((\frac{1}{2}, \frac{11}{2})\) b \((\frac{1}{3}, \frac{14}{3})\) c \((\sqrt{8}, 3 + 10*\sqrt{2})\) d \((\sqrt{4}, 13)\) e \((\sqrt{2}, \frac{31}{3})\)

Even without calculations it is clear that ( \(\sqrt{2}\) , \(\frac{31}{3}\) ) cannot lie on the line \(y = 5x + 3\) . If x is irrational, y must also be irrational.

What do they mean by \((\sqrt{2}, \frac{31}{3})\) is irrational?

agree with E for the reason given above.

\(\sqrt{2}\) is an irrational number i.e a non-terminating value. _________________

\(\sqrt{2}\) is an irrational number i.e a non-terminating value.

Ah, there's a difference between 'irrational' and 'non-terminating'. 1/3 = 0.33333.... is 'non-terminating', but it is also certainly a rational number. Rational numbers are those that can be written as fractions using only integers. As decimals, rational numbers can be terminating or non-terminating, but when rational numbers produce non-terminating decimals, they *always* produce repeating (sometimes called recurring) decimals; a certain pattern of digits repeats forever. Irrational numbers are those numbers like \(\Pi\) and \(\sqrt{2}\) which cannot be written as fractions involving integers. As decimals, irrational numbers have no pattern of digits that repeats forever. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

2\sqrt{2} is irrational too?(and transcendental number too ... ) I did not get the explanation about it being sqrt{2}irrational number ..so not an option _________________

Even without calculations it is clear that ( \(\sqrt{2}\) , \(\frac{31}{3}\) ) cannot lie on the line \(y = 5x + 3\) . If x is irrational, y must also be irrational.

What do they mean by \((\sqrt{2}, \frac{31}{3})\) is irrational?

i am going to follow substitution method...

A: 1/2, 11/2 which is in the form of x,y

if substituting these values in y = 5x + 3 makes LHS = RHS then thats the answer

Even without calculations it is clear that ( \(\sqrt{2}\) , \(\frac{31}{3}\) ) cannot lie on the line \(y = 5x + 3\) . If x is irrational, y must also be irrational.

What do they mean by \((\sqrt{2}, \frac{31}{3})\) is irrational?

i am going to follow substitution method...

A: 1/2, 11/2 which is in the form of x,y

if substituting these values in y = 5x + 3 makes LHS = RHS then thats the answer

\frac{31}{3} = 5 * \sqrt{2} + 3 = 5 * 1.41 + 3 approximately 10 so may be YES

i am getting YES for most of the substitution...am i doing anything wrong?

Guys, please help me understanding this problem...Bunuel please help

Substituting values is not the best way to solve this question but anyway: if you go this way then when you substitute the values of x and y, LHS and RHS must be equal without any approximation.

For \((\sqrt{2}, \frac{31}{3})\) --> \(LHS=y=\frac{31}{3}=rational\) and \(RHS=5x + 3=5\sqrt{2}+3=irrational\) --> \(rational\neq{irrational}\), so line \(y=5x+3\) does not pass through point \((\sqrt{2}, \frac{31}{3})\)

P.S. To format the formulas correctly please mark them and push [ m] button. _________________

Even without calculations it is clear that ( \(\sqrt{2}\) , \(\frac{31}{3}\) ) cannot lie on the line \(y = 5x + 3\) . If x is irrational, y must also be irrational.

What do they mean by \((\sqrt{2}, \frac{31}{3})\) is irrational?

i am going to follow substitution method...

A: 1/2, 11/2 which is in the form of x,y

if substituting these values in y = 5x + 3 makes LHS = RHS then thats the answer

\frac{31}{3} = 5 * \sqrt{2} + 3 = 5 * 1.41 + 3 approximately 10 so may be YES

i am getting YES for most of the substitution...am i doing anything wrong?

Guys, please help me understanding this problem...Bunuel please help

Substituting values is not the best way to solve this question but anyway: if you go this way then when you substitute the values of x and y, LHS and RHS must be equal without any approximation.

For \((\sqrt{2}, \frac{31}{3})\) --> \(LHS=y=\frac{31}{3}=rational\) and \(RHS=5x + 3=5\sqrt{2}+3=irrational\) --> \(rational\neq{irrational}\), so line \(y=5x+3\) does not pass through point \((\sqrt{2}, \frac{31}{3})\)

P.S. To format the formulas correctly please mark them and push [ m] button.

Even without calculations it is clear that ( \(\sqrt{2}\) , \(\frac{31}{3}\) ) cannot lie on the line \(y = 5x + 3\) . If x is irrational, y must also be irrational.

What do they mean by \((\sqrt{2}, \frac{31}{3})\) is irrational?

i am going to follow substitution method...

A: 1/2, 11/2 which is in the form of x,y

if substituting these values in y = 5x + 3 makes LHS = RHS then thats the answer

\frac{31}{3} = 5 * \sqrt{2} + 3 = 5 * 1.41 + 3 approximately 10 so may be YES

i am getting YES for most of the substitution...am i doing anything wrong?

Guys, please help me understanding this problem...Bunuel please help

Substituting values is not the best way to solve this question but anyway: if you go this way then when you substitute the values of x and y, LHS and RHS must be equal without any approximation.

For \((\sqrt{2}, \frac{31}{3})\) --> \(LHS=y=\frac{31}{3}=rational\) and \(RHS=5x + 3=5\sqrt{2}+3=irrational\) --> \(rational\neq{irrational}\), so line \(y=5x+3\) does not pass through point \((\sqrt{2}, \frac{31}{3})\)

P.S. To format the formulas correctly please mark them and push [ m] button.

How are you supposed to know that this is testing rational/irrational. When you look at this problem, what does one think to crack it open? What about for similar problems that ultimately aren't testing rational/irrational?

Even without calculations it is clear that ( \(\sqrt{2}\) , \(\frac{31}{3}\) ) cannot lie on the line \(y = 5x + 3\) . If x is irrational, y must also be irrational.

What do they mean by \((\sqrt{2}, \frac{31}{3})\) is irrational?

i am going to follow substitution method...

A: 1/2, 11/2 which is in the form of x,y

if substituting these values in y = 5x + 3 makes LHS = RHS then thats the answer

11/2 = 5*1/2 + 3 = 5/2 + 3 =(5+6)/3 (5+6)/2 = 11/3 11/2 so NO YES

I dont think the test wants to check ur equation solving powers with this question. So they would try stumping you with the options.

Can you help us in understanding the concept of rational and irrational and also tell us why if rational not equal to irratonal the line wont pass.. Am not able to grasp the concept _________________

"Giving kudos" is a decent way to say "Thanks" and motivate contributors. Please use them, it won't cost you anything

Can you help us in understanding the concept of rational and irrational and also tell us why if rational not equal to irratonal the line wont pass.. Am not able to grasp the concept

Some point (a, b) is on line y=5x+3 means that, when you replace x and y with a and b, the equation should hold.

For \((\sqrt{2}, \frac{31}{3})\) the left hand side is \(\frac{31}{3}=rational\) and the right hand side is \(5\sqrt{2}+3=irrational\). Does LHS equal to RHS? No, \(rational\neq{irrational}\), so line \(y=5x+3\) does not pass through point \((\sqrt{2}, \frac{31}{3})\).